Exercise 2.7-Additional Questions - Chapter 2 - Algebra - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions Solved
Question 1.
Solve for $x^2-7 x^3+8 x^2+8 x-8=0$. given $3-\sqrt{5}$ is a root Solution:
when $3-\sqrt{5}$ is a root, $3+\sqrt{5}$ is the other root.
$
\begin{aligned}
& \text { S.o.r. }=(3-\sqrt{5})+(3+\sqrt{5})=6 \\
& =(3-\sqrt{5})(3+\sqrt{5})=3^2-\sqrt{5}^2 \\
& =9-5=4 \\
&
\end{aligned}
$

The equation is $x^2-6 x+4=0$
Now $x^4-7 x^3+8 x^2+8 x-8=\left(x^2-6 x+4\right)\left(x^2+p x-2\right)$
Equating co-eff of $x$
$
\begin{aligned}
& 12+4 \mathrm{p}=8 \\
& 4 \mathrm{p}=8-12=-4
\end{aligned}
$
So the other factor is $x^2-\mathrm{x}-2$
Now solving $x^2-x-2=0$
$
\begin{aligned}
& x=\frac{1 \pm \sqrt{1-4(1)(2)}}{2}=\frac{1 \pm \sqrt{9}}{2} \\
& x=\frac{1+3}{2}, \frac{1-3}{2} ; x=2,-1
\end{aligned}
$
Question 2.
Solve the equation $\mathrm{x}^3+5 \mathrm{x}^2-16 \mathrm{x}-14=0$. given $\mathrm{x}+7$ is a root
Solution:
$
\mathrm{x}^3+5 \mathrm{x}^2-16 \mathrm{x}-14=(\mathrm{x}+7)\left(\mathrm{x}^2+\mathrm{px}-2\right)
$
Equating co-eff of $\mathrm{x}$
$
\begin{aligned}
& 7 \mathrm{p}-2=-16 \\
& 7 \mathrm{p}=-16+2=-14 \\
& \Rightarrow \mathrm{p}=-2
\end{aligned}
$
So the other factor is $\mathrm{x}^2-2 \mathrm{x}-2$
Solving $x^2-2 x-2=0$